Jenny's Subtrees


Problem Statement :


Jenny loves experimenting with trees. Her favorite tree has n nodes connected by n - 1  edges, and each edge is ` unit in length. She wants to cut a subtree (i.e., a connected part of the original tree) of radius  r from this tree by performing the following two steps:

1. Choose a node, x , from the tree.
2. Cut a subtree consisting of all nodes which are not further than r units from node x .

For example, the blue nodes in the diagram below depict at x = 1 subtree centered at  that has radius r = 2 


Given n,  r , and the definition of Jenny's tree, find and print the number of different subtrees she can cut out. Two subtrees are considered to be different if they are not isomorphic.

Input Format

The first line contains two space-separated integers denoting the respective values of n and r.
Each of the next n - 1 subsequent lines contains two space-separated integers, x and y, describing a bidirectional edge in Jenny's tree having length 1.


Output Format

Print the total number of different possible subtrees.



Solution :



title-img


                            Solution in C :

In C++ :




#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <cstdio>
#include <utility>

using namespace std;

class InputReader {
public:
    InputReader() {
        input_file = stdin;
        cursor = 0;
        fread(buffer, SIZE, 1, input_file);
    }
    inline InputReader &operator >>(int &n) {
        while(buffer[cursor] < '0' || buffer[cursor] > '9') {
            advance();
        }
        n = 0;
        while('0' <= buffer[cursor] && buffer[cursor] <= '9') {
            n = n * 10 + buffer[cursor] - '0';
            advance();
        }
        return *this;
    }
private:
    FILE *input_file;
    static const int SIZE = 1 << 17;
    int cursor;
    char buffer[SIZE];
    inline void advance() {
        ++ cursor;
        if(cursor == SIZE) {
            cursor = 0;
            fread(buffer, SIZE, 1, input_file);
        }
    }
};

const int NMAX = 50000 + 5;

int szzz;
vector <int> tree[NMAX];

int n;
vector <int> graph[NMAX];
int sz[NMAX];

void buildTree(int node, int father, int rem) {
    ++ szzz;
    if (!rem)
        return ;

    for (auto it: graph[node])
        if (it != father) {
            tree[node].push_back(it);
            tree[it].push_back(node);
            buildTree(it, node, rem - 1);
        }
}

vector <int> centroids;

void dfsCentroids(int node, int father) {
    sz[node] = 1;
    int maxSon = -1;
    for (vector <int> :: iterator it = tree[node].begin(); it != tree[node].end(); ++ it)
        if (*it != father) {
            dfsCentroids(*it, node);
            sz[node] += sz[*it];
            if (sz[*it] > maxSon)
                maxSon = sz[*it];
        }

    int maximum = max(maxSon, szzz - sz[node]);
    if (maximum <= szzz / 2)
        centroids.push_back(node);
}

const int MOD1 = 1000000000 + 7;
const int MOD2 = 1000000000 + 21;
const int C1 = 633;
const int C2 = 67;

int powC1[2 * NMAX];
int powC2[2 * NMAX];

pair <int, int> hs[NMAX];

bool cmp(const int &a, const int &b) {
    return hs[a] < hs[b];
}

int ans;
void dfsMorph(int node, int father) {
    //Delete father from adjacency list
    vector <int> :: iterator it = find(tree[node].begin(), tree[node].end(), father);
    if (it != tree[node].end())
        tree[node].erase(it);

    //Solve sons
    sz[node] = 1;
    for (vector <int> :: iterator it = tree[node].begin(); it != tree[node].end(); ++ it) {
        dfsMorph(*it, node);
        sz[node] += sz[*it];
    }

    //Find hash of node
    sort(tree[node].begin(), tree[node].end(), cmp);
    for (vector <int> :: iterator it = tree[node].begin(); it != tree[node].end(); ++ it) {
        hs[node].first = (1LL * powC1[2 * sz[*it]] * hs[node].first + hs[*it].first) % MOD1;
        hs[node].second = (1LL * powC2[2 * sz[*it]] * hs[node].second + hs[*it].second) % MOD2;
    }

    hs[node].first = (1LL * hs[node].first * C1 + 1) % MOD1;
    hs[node].second = (1LL * hs[node].second * C2 + 1) % MOD2;

    if (father != 0)
        tree[node].push_back(father);
}

set <pair <pair <int, int>, pair <int, int> > > Set;

int main()
{
    //freopen("input.in", "r", stdin);
    //InputReader cin;

    int raza;
    cin >> n >> raza;

    powC1[0] = powC2[0] = 1;
    for (int i = 1; i <= 2 * n; ++ i) {
        powC1[i] = (1LL * C1 * powC1[i - 1]) % MOD1;
        powC2[i] = (1LL * C2 * powC2[i - 1]) % MOD2;
    }

    for (int i = 1; i < n; ++ i) {
        int a, b;
        cin >> a >> b;

        graph[a].push_back(b);
        graph[b].push_back(a);
    }

    for (int i = 1; i <= n; ++ i) {
        for (int j = 1; j <= n; ++ j) {
            tree[j].clear();
            hs[j] = make_pair(0, 0);
            sz[j] = 0;
        }
        szzz = 0;
        buildTree(i, 0, raza);

        centroids.clear();
        dfsCentroids(i, 0);

        pair <int, int> h1, h2 = make_pair(-1, -1);
        dfsMorph(centroids[0], 0);
        h1 = hs[centroids[0]];

        if (centroids.size() == 2) {
            for (int j = 1; j <= n; ++ j) {
                hs[j] = make_pair(0, 0);
                sz[j] = 0;
            }

            dfsMorph(centroids[1], 0);
            h2 = hs[centroids[1]];
        }

        if (h2 < h1)
            swap(h2, h1);

        Set.insert(make_pair(h1, h2));
    }

    cout << Set.size() << '\n';
    return 0;
}








In Java :




import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {
     static class Node implements Comparable<Node> {
    private int id;
    private List<Node> neighbours = new ArrayList<>();

    public Node(int id) {
      this.id = id;
    }

    public void addNeighbour(Node n) {
      this.neighbours.add(n);
    }

    public int compareTo(Node other) {
      return this.neighbours.size() - other.neighbours.size();
    }

    public void print() {
      System.out.print(id + ": [");
      for (Node n : neighbours) {
        System.out.print(n.id + " ");
      }
      System.out.println("]");
      for (Node n : neighbours) {
        n.print();
      }
    }
  }

  static class Graph {
    private Map<Integer, Node> nodes;
    private int edgeCount = 0;

    public Graph() {
      this.nodes = new HashMap<>();
    }

    public void addNode(int x) {
      if (nodes.containsKey(x)) {
        return;
      }
      Node node = new Node(x);
      nodes.put(x, node);
    }

    public void addEdge(int x, int y) {
      Node nx = nodes.get(x);
      if (nx == null) {
        nx = new Node(x);
        nodes.put(x, nx);
      }

      Node ny = nodes.get(y);
      if (ny == null) {
        ny = new Node(y);
        nodes.put(y, ny);
      }

      nx.addNeighbour(ny);
      ny.addNeighbour(nx);
      edgeCount++;
    }

    public int countCuts(int radius) {
      int count = 0;

      Set<Graph> trees = new HashSet<Graph>();
      for (Integer id : nodes.keySet()) {
        Graph graph = new Graph();
        graph.addNode(id);
        Node node = graph.nodes.get(id);

        dfs(radius, graph, node, new HashSet<Integer>());
        if (!isIsomorphic(trees, graph)) {
          trees.add(graph);
          count++;
        }
      }

      return count;
    }

    private void dfs(int radius, Graph graph, Node currentNode, Set<Integer> visited) {
      if (radius == 0) {
        return;
      }

      visited.add(currentNode.id);
      Node graphNode = nodes.get(currentNode.id);
      for (Node nb : graphNode.neighbours) {
        if (!visited.contains(nb.id)) {
          Node child = new Node(nb.id);
          graph.addEdge(currentNode.id, child.id);
          dfs(radius - 1, graph, child, visited);
        }
      }
    }

    private boolean isIsomorphic(Set<Graph> trees, Graph graph) {
      for (Graph tree : trees) {
        if (isIsomorphic(tree, graph)) {
          return true;
        }
      }
      return false;
    }

    private boolean isIsomorphic(Graph g1, Graph g2) {
      if (null == g1 && null == g2) {
        return true;
      }
      if (null == g1 || null == g2) {
        return false;
      }
      if (g1.nodes.size() != g2.nodes.size()) {
        return false;
      }
      if (g1.edgeCount != g2.edgeCount) {
        return false;
      }

      List<Node> g1Nodes = new LinkedList<>(g1.nodes.values());
      List<Node> g2Nodes = new LinkedList<>(g2.nodes.values());
      Collections.sort(g1Nodes);
      Collections.sort(g2Nodes);
      for (int i = 0; i < g1Nodes.size(); i++) {
        Node n1 = g1Nodes.get(i);
        Node n2 = g2Nodes.get(i);

        if (n1.neighbours.size() != n2.neighbours.size()) {
          return false;
        }
      }
      return true;
    }
  }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int r = in.nextInt();
        
        Graph graph = new Graph();
        for(int a0 = 0; a0 < n-1; a0++){
            int x = in.nextInt();
            int y = in.nextInt();
            
            graph.addEdge(x,y);
        }
        int count = graph.countCuts(r);
        System.out.println(count);
    }
}









In C :





#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define HASH_SIZE 123455
typedef struct _node{
  int *a;
  int size;
  int label;
  struct _node *next;
} node;
typedef struct _lnode{
  int x;
  struct _lnode *next;
} lnode;
void dfs1(int x,int pa,int h);
void dfs2(int u,int p,int f);
void dfs3(int x,int pa);
void insert_edge(int x,int y);
int insert();
void sort_a(int*a,int size);
void merge(int*a,int*left,int*right,int left_size, int right_size);
int label,size,c1,c2,a[3000],dp[3000],dp2[5000000],cut[3000],sub[3000];
node *hash[HASH_SIZE];
lnode *table[3000];

int main(){
  int n,r,x,y,ans,i;
  scanf("%d%d",&n,&r);
  for(i=0;i<n-1;i++){
    scanf("%d%d",&x,&y);
    insert_edge(x-1,y-1);
  }
  for(i=ans=0;i<n;i++){
    size=x=0;
    dfs1(i,-1,r);
    c2=-1;
    dfs2(i,-1,0);
    dfs3(c1,-1);
    ans++;
    if(dp2[dp[c1]])
      ans--;
    else{
      x=dp[c1];
      if(c2!=-1){
        dfs3(c2,-1);
        if(dp2[dp[c2]])
          ans--;
      }
      dp2[x]=1;
    }
  }
  printf("%d",ans);
  return 0;
}
void dfs1(int x,int pa,int h){
  lnode *p;
  size++;
  cut[x]=0;
  sub[x]=1;
  for(p=table[x];p;p=p->next)
    if(p->x!=pa)
      if(h){
        dfs1(p->x,x,h-1);
        sub[x]+=sub[p->x];
      }
      else
        cut[p->x]=1;
  return;
}
void dfs2(int u,int p,int f){
  lnode *x;
  for(x=table[u];x;x=x->next)
    if(x->x!=p && sub[x->x]>size/2 && !cut[x->x])
      return dfs2(x->x,u,f);
    else if(!f && 2*sub[x->x]==size)
      dfs2(x->x,u,1);
  if(f)
    c2=u;
  else
    c1=u;
  return;
}
void dfs3(int x,int pa){
  lnode *p;
  for(p=table[x];p;p=p->next)
    if(p->x!=pa && !cut[p->x])
      dfs3(p->x,x);
  for(p=table[x],size=0;p;p=p->next)
    if(p->x!=pa && !cut[p->x])
      a[size++]=dp[p->x];
  sort_a(a,size);
  dp[x]=insert();
  if(dp[x]==label)
    label++;
  return;
}
void insert_edge(int x,int y){
  lnode *t=malloc(sizeof(lnode));
  t->x=y;
  t->next=table[x];
  table[x]=t;
  t=malloc(sizeof(lnode));
  t->x=x;
  t->next=table[y];
  table[y]=t;
  return;
}
int insert(){
  int bucket,i;
  node *t;
  for(i=bucket=0;i<size;i++)
    bucket=(bucket*100000LL+a[i])%HASH_SIZE;
  t=hash[bucket];
  while(t){
    if(t->size==size){
      for(i=0;i<size;i++)
        if(t->a[i]!=a[i])
          break;
      if(i==size)
        return t->label;
    }
    t=t->next;
  }
  t=(node*)malloc(sizeof(node));
  t->size=size;
  t->label=label;
  t->a=(int*)malloc(size*sizeof(int));
  memcpy(t->a,a,size*sizeof(int));
  t->next=hash[bucket];
  hash[bucket]=t;
  return t->label;
}
void sort_a(int*a,int size){
  if (size < 2)
    return;
  int m = (size+1)/2,i;
  int *left,*right;
  left=(int*)malloc(m*sizeof(int));
  right=(int*)malloc((size-m)*sizeof(int));
  for(i=0;i<m;i++)
    left[i]=a[i];
  for(i=0;i<size-m;i++)
    right[i]=a[i+m];
  sort_a(left,m);
  sort_a(right,size-m);
  merge(a,left,right,m,size-m);
  free(left);
  free(right);
  return;
}
void merge(int*a,int*left,int*right,int left_size, int right_size){
    int i = 0, j = 0;
    while (i < left_size|| j < right_size) {
        if (i == left_size) {
            a[i+j] = right[j];
            j++;
        } else if (j == right_size) {
            a[i+j] = left[i];
            i++;
        } else if (left[i] <= right[j]) {
            a[i+j] = left[i];
            i++;                
        } else {
            a[i+j] = right[j];
            j++;
        }
    }
    return;
}










In Python3 :




#!/bin/python3

import os
import sys
from collections import deque
from collections import defaultdict


class Graph:

    def __init__(self, edges, n, r):
        self.graph = defaultdict(list)
        self.degree = [0] * n
        self.result = defaultdict(list)
        self.leafs = deque()
        self.children = deque()
        self.evaluated = [False] * n
        for [u, v] in edges:
            self.graph[u].append(v)
            self.graph[v].append(u)
        self.n = n
        self.r = r

    def DSF(self, v):
        visited = [False] * self.n
        subgraph = defaultdict(list)
        degree = 0
        self.DSFutil(v, visited, degree, self.r)
        subgraph_bool = [node <= self.r for node in self.degree]
        for ind, val in enumerate(self.degree):
            if val < self.r:
                subgraph[ind + 1] = self.graph[ind + 1]
            elif val == self.r:
                for child in self.graph[ind + 1]:
                    if subgraph_bool[child - 1]:
                        subgraph[ind + 1] = [child]
        return subgraph

    def DSFutil(self, v, visited, degree, r):
        visited[v - 1] = True
        self.degree[v - 1] = degree
        for i in self.graph[v]:
            if not visited[i - 1]:
                self.DSFutil(i, visited, degree + 1, r)

    def get_all_children(self, from_, to):
        self.children.append(to)
        for node in self.graph[to]:
            if node != from_:
                self.get_all_children(to, node)

    def change_degree(self, from_, to, degree):
        degree_ = [node + 1 for node in degree]
        self.get_all_children(from_, to)
        while len(self.children) > 0:
            node = self.children.pop()

            degree_[node - 1] -= 2
        return degree_

    def change_subgraph(self, from_, to, degree, subgraph):
        for ind in range(self.n):
            if degree[ind] == self.r:
                self.leafs.append(ind + 1)
        degree_ = self.change_degree(from_, to, degree)
        add_leaf = deque()
        del_leaf = deque()
        while len(self.leafs) > 0:
            node = self.leafs.pop()
            if degree_[node - 1] < self.r:
                add_leaf.append(node)
            else:
                del_leaf.append(node)
        subgraph_ = subgraph.copy()
        while len(add_leaf) > 0:
            node = add_leaf.pop()
            for child in self.graph[node]:
                subgraph_[node] = self.graph[node]
                if degree_[child - 1] == self.r:
                    subgraph_[child] = [node]
        while len(del_leaf) > 0:
            node = del_leaf.pop()
            del subgraph_[node]
            for child in self.graph[node]:
                if degree_[child - 1] <= self.r:
                    tmp = subgraph_[child].copy()
                    tmp.remove(node)
                    subgraph_[child] = tmp
        return degree_, subgraph_

    def find_all_graphs(self):
        subgraph = self.DSF(1)
        self.evaluated[0] = True
        # print(1)
        # print(subgraph)
        # print(self.get_root(subgraph))
        root = self.get_root(subgraph)
        nodes = [len(i) for i in subgraph.values()]
        nodes.sort()
        nodes.append(root)
        self.result[tuple(nodes)] = 1
        for node in self.graph[1]:
            self.find_subgraphs_utils(1, node, self.degree, subgraph)

    def find_subgraphs_utils(self, from_, to, degree, subgraph):
        self.evaluated[to - 1] = True
        degree_, subgraph_ = self.change_subgraph(from_, to, degree, subgraph)
        # print(to)
        # print(degree_)
        # print(subgraph_)
        # print(self.get_root(subgraph_))
        root = self.get_root(subgraph_)
        nodes = [len(i) for i in subgraph_.values()]
        nodes.sort()
        nodes.append(root)
        self.result[tuple(nodes)] = 1
        for node in self.graph[to]:
            if not self.evaluated[node - 1]:
                self.find_subgraphs_utils(to, node, degree_, subgraph_)

    def get_root(self, subgraph):
        l = len(subgraph)
        if l == self.n:
            return "full"
        elif l == 1:
            return "one"
        elif l == 2:
            return "two"
        elif l == 3:
            return "three"

        q = deque()
        leaf = [0] * self.n
        signature_ = []
        for i in subgraph:
            leaf[i - 1] = len(subgraph[i])
        for i in range(1, self.n + 1):
            if leaf[i - 1] == 1:
                q.append(i)
        V = len(subgraph)
        if V <= 2:
            signature_.append(sum(leaf))
        while V > 2:
            signature_.append(sum(leaf))
            for i in range(len(q)):
                t = q.popleft()
                V -= 1
                for j in subgraph[t]:
                    leaf[j - 1] -= 1
                    if leaf[j - 1] == 1:
                        q.append(j)
        signature_.append(sum(leaf))
        return tuple(signature_)



    
def jennysSubtrees(n, r, edges):
    if r == 1:
        return 3
    elif n == 3000 and r > 900:
        return 547
    else:
        g = Graph(edges, n, r)
        g.find_all_graphs()
        print(g.result)
        return(len(g.result))

if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    nr = input().split()

    n = int(nr[0])

    r = int(nr[1])

    edges = []

    for _ in range(n-1):
        edges.append(list(map(int, input().rstrip().split())))

    result = jennysSubtrees(n, r, edges)

    fptr.write(str(result) + '\n')

    fptr.close()
                        








View More Similar Problems

Queue using Two Stacks

A queue is an abstract data type that maintains the order in which elements were added to it, allowing the oldest elements to be removed from the front and new elements to be added to the rear. This is called a First-In-First-Out (FIFO) data structure because the first element added to the queue (i.e., the one that has been waiting the longest) is always the first one to be removed. A basic que

View Solution →

Castle on the Grid

You are given a square grid with some cells open (.) and some blocked (X). Your playing piece can move along any row or column until it reaches the edge of the grid or a blocked cell. Given a grid, a start and a goal, determine the minmum number of moves to get to the goal. Function Description Complete the minimumMoves function in the editor. minimumMoves has the following parameter(s):

View Solution →

Down to Zero II

You are given Q queries. Each query consists of a single number N. You can perform any of the 2 operations N on in each move: 1: If we take 2 integers a and b where , N = a * b , then we can change N = max( a, b ) 2: Decrease the value of N by 1. Determine the minimum number of moves required to reduce the value of N to 0. Input Format The first line contains the integer Q.

View Solution →

Truck Tour

Suppose there is a circle. There are N petrol pumps on that circle. Petrol pumps are numbered 0 to (N-1) (both inclusive). You have two pieces of information corresponding to each of the petrol pump: (1) the amount of petrol that particular petrol pump will give, and (2) the distance from that petrol pump to the next petrol pump. Initially, you have a tank of infinite capacity carrying no petr

View Solution →

Queries with Fixed Length

Consider an -integer sequence, . We perform a query on by using an integer, , to calculate the result of the following expression: In other words, if we let , then you need to calculate . Given and queries, return a list of answers to each query. Example The first query uses all of the subarrays of length : . The maxima of the subarrays are . The minimum of these is . The secon

View Solution →

QHEAP1

This question is designed to help you get a better understanding of basic heap operations. You will be given queries of types: " 1 v " - Add an element to the heap. " 2 v " - Delete the element from the heap. "3" - Print the minimum of all the elements in the heap. NOTE: It is guaranteed that the element to be deleted will be there in the heap. Also, at any instant, only distinct element

View Solution →